can you get c'thun for free when you buy wotog packs?

now that there is a curropt garrosh portrait i kinda wanted to play c'thun warrior but i don't have c'thun. Now i remember back in the day you would get c'thun and beckoner of evil for free when you opened you first wotog pack. 

My question is if anyone knows if it still applies today. so if i buy let's say 15 wotog packs do i get c'thun and beckoner of evil for free or not?

I never openend a wotog pack on the account i am currently playing on.

thank you in advance

I was pretty sure that was a time-limited offer, but according to the HS wiki to this day, the first WotOG pack you crack open has a C'thun in it.

That said, the power of the card has petered off quite a bit in comparison with what's possible in 2019 hearthstone. C'thun is slow and a bit underwhelming.

I made a new account a couple of months ago and my first Old Gods pack yeilded a C'thun. Mind you this was about 2 years after the Old Gods expansion, so I am pretty sure that rule will be there forever.

answer is yes. you're welcome.

thank all of you for responding. It means alot to me. I guess i'll be buying wotog packs

A little advice to save you some money: if you want to build wild decks, crafting the cards is allmost allways the best choice. Except for the very first pack of wotog, aiming for specific epic or legendary cards is highly luck dependent and expencive.

Now that i think of it, wotog may be an exception because of the high amount of common c‘thun support cards. But still, i wouldn‘t buy more than 10.

Instead, buy the cheaper bundles if you are ready to spend money. It‘s a better dust value and the cards may open up possibilies in standard play aswell. 

Little justification. Standard players buy packs because:

1. Pre purchase is a relativly good deal

2. A hughe chunk of the cards will be playable in the upcoming meta.

Both points aren‘t the case for wild. Meta defining cards are spread over a large number of expansions and can’t reasonably be collectes through card opening.